proof: Limit of self-super-roots is e^1/e. TPID 6 - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: proof: Limit of self-super-roots is e^1/e. TPID 6 (/showthread.php?tid=362) proof: Limit of self-super-roots is e^1/e. TPID 6 - bo198214 - 10/07/2009 In reply to http://math.eretrandre.org/tetrationforum/showthread.php?tid=162&pid=4073#pid4073 First it is easy to see that for $1: ${^n b}=\exp_b^{\circ n}(1)\to a ($a$ is the lower fixed point of $b^x$) Hence for $n_0 > 3$ we have for all $n\ge n_0$: (*) ${^n b} < e < n$ We also know that for $b>\eta$, $\exp_b^{\circ n}\to\infty$ quite fast, particularly for each $b>\eta$ there is an $n_0$ such that for all $n\ge n_0$: (**) ${^n b} > n$. Now we lead proof by contradiction, suppose that $\lim_{n\to\infty} b_n \neq \eta$ where ${^n b_n} = n,\quad b_n > 1$. Then there must be a subsequence $b_{m},\quad m\in M\subseteq\mathbb{N}$ and $\epsilon>0$ such that this subsequence stays always more than $\eps$ apart from $\eta$: $\left|b_{m}-\eta\right| \ge \epsilon$. I.e. there is $B_1<\eta$ and $B_2>\eta$ such that either $b_m \le B_1$ or $b_m \ge B_2$. By (*) and (**) we have $n_0$ such that for all $m\ge n_0$: ${^m B_1} and ${^m B_2}>m$. As ${^m x}$ is monotone increasing for $x>1$ we have also ${^m b_m} and ${^m b_m}>m$. This particularly means ${^m b_m}\neq m$ and hence none of the $b_m$ can be the self superroot, in contradiction to our assumption. RE: proof: Limit of self-super-roots is e^1/e. TPID 6 - andydude - 10/07/2009 Wow! Very nice! You make it seem so easy. I've been working on that one for while, ever since the xsrtx thread. RE: proof: Limit of self-super-roots is e^1/e. TPID 6 - Base-Acid Tetration - 07/10/2010 The same method of proof could possibly be used to easily prove that, possibly for all k>4, limit of self-hyper-k-root(x) as x -> infinity = $\eta_k$ (defined as the largest real x such that $x[k]\infty < \infty$, i.e. where the maximum of self-hyper-(k-1)-root function occurs; let's establish this notation); yeah I know, I only substituted the pentation-analogues into the proof and quickly checked. RE: proof: Limit of self-super-roots is e^1/e. TPID 6 - bo198214 - 07/10/2010 (07/10/2010, 05:19 AM)Base-Acid Tetration Wrote: The same method of proof could possibly be used to easily prove that, possibly for all k>4 The thing is: to define the hyper k-self-root you need a hyper (k-1) operation defined on the reals. And we still have several methods of doing this without equality proofs.